solution manual for Signals and Systems A Primer with MATLAB® 2nd edition by Sadiku

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solution manual for Signals and Systems A Primer with
MATLAB® 2nd edition by Sadiku




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Table of Contents
Chapter 1 1
Chapter 2 41
Chapter 3 77
Chapter 4 117
Chapter 5 144
Chapter 6 182
Chapter 7 205




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CHAPTER 1

P. P. 1.1
2 2
The period is T = = =1
 2
x(t + T ) = A cos((2 (t +1) + 0.1 )
= A cos(2 t + 2 + 0.1 )
= A cos(2 t + 0.1 )
= x(t)
Hence x(t) is periodic.

P.P. 1.2

(a) x(t) = t, 0 < t < 
T /2 T /2  T / 2 3
E = lim
T →
 T →

| x(t) |2dt = limt 2
dt = lim 2   =
−T / 2 −T / 2 T →  3 

1
 | x(t) |2dt = lim 1  t 2dt = lim 2  T / 2  = 
T / 2 T / 2 3
P = lim
T → T T → T T → T  3 
−T / 2 −T / 2



i.e. x(t) is neither an energy nor a power signal.

(b)
T /2 a
E = lim  | x(t) |2dt = lim  A dt = 2aA
2 2
T → T →
−T / 2 −a


i.e. x(t) is an energy signal.

(c) | x(n) |= 5 | e− j4n |= 5
 | x[n] |2 = lim  52
N N
1 1
P = lim
N → 2N +1 N → 2N +1
n=− N n=− N
1
= lim 25(2N +1) = 25 
N → 2N +1



i.e x[n] is a power signal.

P.P. 1.3
(a) ze = t 2 −10, zo = 4t




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1
h (t) = [u(t +1) −u(t −1)]
8s

s8 s8 s8 s8 s8 s8 s8

e 2
(b)
1
h (t) = [−u(t +1) + 2u(t) −u(t −1)]
s8

s 8 s8 s8 s8 s8 s8 s8 8s s8

o
2

These are sketched below. s8 s8 s8 ho
he

½ 1/2



t -1 0 1 t
-1 0 1
-1/2


P. P. 1.4
s8 s8



= sin( / 2) =1
(a)
−
sin(t3 + /2)(t)dt =sin(t3 + /2)
s8
s8 8s s8 s8
8s
s8 s8 s8 s8 8s s8 s8

t =0 s8 s8
s8 s8 s8 s8 8s




10

=1 + 4 − 2 = 3

(b) (t2 + 4t − 2)(t −1)dt = (t2 + 4t − 2)
0
8s
8s s8 s8 s8 s8 s8 s8 s8 s8 8s s8 s8 s8 s8

t =1s8 8s
8s s8 8s s8 s8 s8




P. P. 1.5
s8 s8



 0, t 0
 s8 s8




i(t) =  10, s8 s8
s8
0t  2s8 s8 s8 s8




 −10, 2t  4 s8 s8 s8 s8




i(t) =10u(t) −u(t − 2)−10u(t − 2) −u(t − 4)
s8 8s s8 s8 s8 s8 8s s8 s8 s8 s8 s8 s8




=10[u(t)− 2u(t − 2) +u(t − 4)]
8s s8 s8 s8 s8 s8 s8 s8 s8




t

Let I = s8 s 8


−
idt s8




For t < 0, s 8 s8 s8 s 8 I = 0. s8 s8


t

For 0 < t < 2, I = 10dt =10t
s8 s8 s 8 s8 s8 s8 s8 s8 8s


0
2 t
t
For 2 < t < 4,
s8 s8 s8 s8 s8
I = 10dt − 10dt = 20−10t
s8 s8 s8 s8 s8 s8 8s = 40−10t
s8 8s


0 2
2
4
For t > 4, s 8 s8 s8
I = 20−10ts8 s8 s8 =0 s8


2
Thus,
s8




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